# How do you evaluate the definite integral by the limit definition given int 6dx from [4,10]?

Dec 20, 2016

$36$

#### Explanation:

Geometric Approach

This is the equivalent of drawing a rectangle whose boundaries are the line $y = 6$, $y = 0$, $x = 4$ and $x = 10$.

This rectangle would therefore have a length of $6$ and a height of $6$, which would have an area of $36$ square units.

Since the definite integral is in fact a measure of area, this is the answer.

Using the limit definition

The limit definition of a definite integral is ${\int}_{a}^{b} f \left(x\right) \mathrm{dx} = {\lim}_{n \to \infty} {\sum}_{i = 1}^{n} f \left({c}_{i}\right) \Delta {x}_{i}$.

The parameter $\Delta {x}_{i}$ is given by $\frac{b - a}{n}$. So, here we would have $\frac{10 - 4}{n} = \frac{6}{n}$.

${c}_{i}$ is given by $a + i \Delta {x}_{i} = 4 + \frac{6 i}{n}$.

Therefore:

${\int}_{4}^{10} 6 \mathrm{dx} = {\lim}_{n \to \infty} {\sum}_{i = 1}^{n} f \left({c}_{i}\right) \Delta {x}_{i}$

$= {\lim}_{n \to \infty} {\sum}_{i = 1}^{n} f \left(4 + \frac{6 i}{n}\right) \times \frac{6}{n}$

Since $f \left(4 + \frac{6 i}{n}\right) = 6$:

$= {\lim}_{n \to \infty} {\sum}_{i = 1}^{n} 6 \times \frac{6}{n}$

$= {\lim}_{n \to \infty} {\sum}_{i = 1}^{n} \frac{36}{n}$

Use the formula ${\sum}_{i = 1}^{n} = n$.

$= {\lim}_{n \to \infty} n \frac{36}{n}$

$= {\lim}_{n \to \infty} 36$

$= 36$

Calculus Approach

We integrate using the formula $\int {x}^{n} \mathrm{dx} = \frac{{x}^{n + 1}}{n + 1} + C$.

$\int \left(6\right) \mathrm{dx} = 6 \frac{{x}^{0 + 1}}{0 + 1} = 6 x$

We now rewrite in proper notation, and evaluate using the second fundamental theorem of calculus, which is that ${\int}_{a}^{b} \left(F \left(x\right)\right) = f \left(b\right) - f \left(a\right)$, where $f \left(x\right)$ is the antiderivative of $F \left(x\right)$:

${\left[6 x\right]}_{4}^{10} = 6 \left(10\right) - 6 \left(4\right) = 60 - 24 = 36$

Same answer as before, just using a different method.

Hopefully this helps!